Sửa đề : a, \(x\left(x-2\right)^2-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x\left(x^2-4x+4\right)-\left(x^3-8\right)=x^3-4x^2+4x-x^3+8\)

\(=-4x^2+4x+8\)

x( x - 2 )² - ( x - 2 )( x² + 2x + 4 ) < bỏ cái ² đi nhá :v >

= x( x² - 4x + 4 ) - ( x³ - 8 )

= x³ - 4x² + 4x - x³ + 8

= -4x² + 4x + 8

a) \(x^{m+2}-2x^m=x^m\left(x^2-2\right)\)

b) \(x^{k+1}-x^{k+2}=x^{k+1}\left(1-x\right)\)

a) x^m+2 - 2x^m = x^m.x² + 2.x^m = x^m( x² - 2 ) = x^m( x - √2 )( x + √2 )

b) x^k+1 - x^k+2 = x^k+1 - x^k+1.x = x^k+1( 1 - x )

PTĐTTNT ??? :)) bn phân tích rồi đấy, đề là tìm x thôi 

Giải ( suỵt :), đừng ai nhìn thấy ... :v 

\(\left(2x-10\right)\left(x+10\right)\left(x+\sqrt{3}\right)=0\)

TH1 : \(2x-10=0\Leftrightarrow x=5\)

TH2 : \(x+10=0\Leftrightarrow x=-10\)

TH3 : \(x+\sqrt{3}=0\Leftrightarrow x=-\sqrt{3}\)( vô lí )

Vậy x = {5;-10}

sao lại "vô lí" vậy bạn 

\(x^3+9x^2+26x+24=\left(x^2+7x+12\right)\left(x+2\right)=\left(x+3\right)\left(x+4\right)\left(x+2\right)\)

Ta có: \(x^3+9x^2+26x+24\)

\(=\left(x^3+2x^2\right)+\left(7x^2+14x\right)+\left(12x+24\right)\)

\(=x^2\left(x+2\right)+7x\left(x+2\right)+12\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+7x+12\right)\)

\(=\left(x+2\right)\left[\left(x^2+3x\right)+\left(4x+12\right)\right]\)

\(=\left(x+2\right)\left[x\left(x+3\right)+4\left(x+3\right)\right]\)

\(=\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

Đến đây là PT tích r còn gì, \(x\in\left\{5;-10;-\sqrt{3}\right\}\)

:v ez mà :) Sửa đề : tìm x 

\(\left(2x-10\right)\left(x+10\right)\left(x+\sqrt{3}\right)=0\)

TH1 : \(2x-10=0\Leftrightarrow x=5\)

TH2 : \(x+10=0\Leftrightarrow x=-10\)

TH3 : \(x+\sqrt{3}=0\Leftrightarrow x=-\sqrt{3}\)( vô lí )

Vậy x = 5 ; x = -10 

Hóa trị của Nito là 2,3,4 bạn nhé .

\(x^2-8x+15\)

\(=x^2-8x+16-1\)

\(=\left(x-4\right)^2-1\)

\(=\left(x-4-1\right)\left(x-4+1\right)\)

\(x^2-8x+15=x^2-8x+16-1=\left(x-4\right)^2-1^2\)

\(=\left(x-4-1\right)\left(x-4+1\right)=\left(x-5\right)\left(x-3\right)\)

Quan sát kĩ sẽ thấy dạng bình phương thiếu nhá ! 

\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=x^3+8-x^3-2x=8-2x\)

( x + 2 )( x² - 2x + 4 ) - x( x² + 2 )

= x³ + 8 - x³ - 2x

= 8 - 2x

a, \(x^3-2x^2+3x-6=x\left(x^2+3\right)-2\left(x^2+3\right)=\left(x-2\right)\left(x^2+3\right)\)

b, \(x^2+2x+1-4y^2=\left(x+1\right)^2-\left(2y\right)^2=\left(x+1-2y\right)\left(x+1+2y\right)\)

\(\left(-2x\right)\left(3x+1\right)+\left(x-2\right)\left(2x+1\right)=-6x^2-2x+2x^2+x-4x-2\)

\(=-4x^2-5x-2\)

Sửa 2x + 1 => 3x + 1 có vẻ sẽ ok hơn nhé ! 

a,\(\left(x-\frac{1}{3}\right)\left(2-x\right)\)

\(=x\left(2-x\right)-\frac{1}{3}\left(2-x\right)\)

\(=2x-x^2-\frac{2}{3}+\frac{1}{3}x\)

\(=x^2+\frac{7}{3}x-\frac{2}{3}\)

b,\(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x^2\)

\(=\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)-3x^2\)

\(=-3x^2\)

a) ( x - 1/3 )( 2 - x ) = x( 2 - x ) -1/3( 2 - x ) = 2x - x² - 2/3 + 1/3x = -x² + 7/3x - 2/3

b) ( x - 1 )³ - ( x - 1 )( x² + x + 1 ) - 3x²

= x³ - 3x² + 3x - 1 - ( x³ - 1 ) - 3x²

= x³ - 6x² + 3x - 1 - x³ + 1

= -6x² + 3x